Question
Round to two sig figs !

323.0 °C | 6.0 × 10 204.0 °C 13.6 1010 Assuming the rate constant obeys the Arrhenius equation, calculate the activation energy E for this reaction. Round your answer to 2 significant digits. kT mol
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Answer #1

ln(K2/k1) = Ea/R[1/T1 - 1/T2]

k1 = 3.6*10^10 , T1 = 204+273.15 = 477.15 k

k2 = 6.0*10^10 , T2 = 323.0+273.15 = 596.15 k
Ea = x kj/mol , R = 8.314*10^-3 j.k-1.mol-1

ln((6*10^10)/(3.6*10^10)) = ((x*10^3)/(8.314))((1/477.15)-(1/596.15)

X = Ea = activation energy = 10 kj/mol

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